Answer :

Given:


Area of the virtual image of each square, A = 6.25 mm2


Area of each square, A0 = 1 mm2


Hence, the linear magnification of the object can be calculated as:


…(1)


Where, m = magnification


A = Area of virtual image


A0 = Area of each square


Putting values in equation (1) we get,


m =


m = 2.5


Also,


m = Image distance(v)/Object distance(u)


m = v/u


Therefore, v = m × u


= 2.5u …(2)


The magnifying glass has a focal length of, f = 10 cm


Applying the lens formula for lens we have:


…(3)


Where, f0 = focal length of the objective lens


v0 = Distance of image formation


u0 = Distance of object from objective


Putting values in the equation (3) we get,



u = -(1.5 × 10) / 2.5


Therefore, u = -6 cm


And v = 2.5u


v = 2.5 × 6


v = -15 cm


The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.


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